meme

I've been having trouble with a lab report that was due a couple of days ago which I still haven't been able to figure out, so I was hoping someone here could please kindly help me (I'd REALLY appreciate it). Its on the equilibrium constant.

Description of the experiment:One set of solutions having known molar concentrations of FeNCS2+ is prepared for a standardization curve, a plot of absorbance versus concentration. A second set of solutions is prepared and mixed to determine the molar concentration of FeNCS2+. By carefully measuring the initial amounts of reactants placed in the reactin systems and the absorbance, the mass action expression at equilibrium can be solved; this equals Kc

Okay, so I've done the first part of the experiment, and I already have my standardization curve. It's the second part of the experiment where I get mixed up.

These is one of the solutions:

Solution # 0.002 M Fe(NO3)3 0.002 M NaSCN 0.1 M HNO35 5mL 1mL 4mL

The blanks we had to fill in on our report sheet:Volume of Fe(NO3)3 (mL) - 5Moles of Fe3+, initial: (calculated this to be: 1.0 x 10-5Volume of NaSCN (mL): 1Moles of SCN-, initial: (calculated this to be: 2.0 x 10-6From the spectrophotometer:Percent transmittance, %T: 82.6Absorbance, A: 0.083

From the calibration curve we got in the first part of the experiment: [ FeNCS2+ ] was: 0.17

Hope you don't mind, I fixed a superscript problem in your post. Thank you, BTW, for using superscripts and subscripts in your post. Most people don't use them and it makes it a pain to read posts without them.

OK, the first thing to do is take a look at the chemical reaction that is in equilibrium here:

Fe+3 + SCN- --> FeSCN+2

Now, I'm assuming that you've learned about the ICE charts. They are charts used to describe the initial concentration, the change in concentration, and the equilibrium concentration of all species present in solution. You fill in the appropriate data. From what you have given in your problem, the chart would look like this:

Fe+3 + SCN- --> FeSCN+2I: 1.0 x 10-5 2.0 x 10-6 ?C: E: 0.17

That's the data you've given. Now, are you sure your calibration curve isn't in units of 10-5 M or something like this? Because there's no way to get 0.17 moles from 0.00005 moles to begin with. I used to TA a lab that was almost the exact same thing, and people got hung up on that all the time.

Now, the change part is where you add variables. You know that the iron III and the thiocyanate are going to come together to form the iron thiocyanate, but you're not sure to what extent. And you know that for every mol of iron and thiocyanate that react, you get one mol of iron thiocyanate (just stoichiometry). This means that for every mol of iron that disappears on the left side of the equation, a mol of iron thiocyanate will appear on the right side. The same goes for thiocyanate. We arbitarily say that this change in mols is the variable X. So, we will in our ICE chart some more (and correct our concentration of FeSCN+2:

Why did we put negative X on the left? That's because you start with only reactants, and the reaction must shift to the right in order to obtain equilibrium. Remember, this is the change that must occur to reach equilibrium. So you dump the two reactants together, and their concentration goes down as the concentration of the products goes up. Does that make sense?

OK, so, now you're asking, 'How do we find out the equilibrium concentrations of the iron and the thiocyanate?' Well, you can write them in terms of X as well.

Now you can write an expression for K. K is given as concentration of products over concentration of reactants at equilibrium, and so it looks like this:

[ FeSCN+2 ] Kc = [ Fe+3 ][ SCN- ]

From your table, you can fill these values in:

[ 1.7 x 10 -6 ] Kc = [ 1.0 x 10-5 - X ][ 2.0 x 10-6 - X ]

Take a look at the ICE chart. Is there anywhere for the FeSCN+2 value to have come from other than the X? Nope. This means that the equilibrium mols of the iron thiocyanate is X. Now that you know that, this whole problem just becomes math. You can calculate all that you asked for, plus the equilibrium constant.

The only other thing to watch out for is that problem is asking for equilibrium concentrations, so once you get your # of mols, you'll have to divide by the number of milliliters of solution that you have in order to calculate the molarity of the chemicals. Take a stab at it and see how it turns out. Come on back here if you have any futher questions.

meme

Thanks for your help, but I still don't understand how to do the math part of the equation. You said to use:

[ 1.7 x 10-6 ]Kc = [ 1.0 x 10-5 - X ][ 2.0 x 10-6 - X ]

Kc and X are both unknowns. I can't solve the equation without knowing the values for either. In our lab manual, Kc is the last value you obtain. You're supposed to have the concentrations of all the reactants and products and just divide it to obtain Kc. Thats why we have all those other things we need to solve for.

Anyway.. the numbers you put into the equation are the number of the moles of each. As I understand it, you're only supposed to use concentrations of the solutions there, so.. M = m/V. (I'm not sure whether you're supposed to include the HNO3 in the volume since it didnt' take any part in the reaction.) I just ended up dividing each by .01 and got:

1.7 x 10-4 M = [FeSCN+2 ]1.0 x 10-3 M = [Fe+3 ]2.0 x 10-4 M = [SCN-]

I tried attaching my graph so you could see it, but I don't think the board takes Excel or Word documents.

"Take a look at the ICE chart. Is there anywhere for the FeSCN+2 value to have come from other than the X? Nope. This means that the equilibrium mols of the iron thiocyanate is X. Now that you know that, this whole problem just becomes math. You can calculate all that you asked for, plus the equilibrium constant."

That should tell you that X = 1.7 x 10-4 M. And, you are correct, you should only be using concentrations, not mols, as I erroniously did on my first time through (hey, I posted at 3:27am, give me a break).

So, X is not an unknown, which means you can solve for Kc.

Also, don't worry about the HNO3. It's just there to solvate the Fe+3; however, you do need to be careful to include it in your volume calculations, as it does affect the volume and thus the concentration.

Unless you're evaporating a lot of the solution down, it's way too big. If you start with .5 mols of each reactant and they react in a 1:1 ratio, how can you get 20 mols back? In your case, you've got two concentrations that are abysmally small and you're getting something back that is nearly 4 orders of magnitude that size.